MHB Solve Christmas Light Wiring Problem: Series vs Parallel

[Dustinsfl]

I was asked the following question and had no idea how to solve it.

If we are concerned with cost, should we wire our christmas lights in series or parallel? Note you can also wire them in a parallel series. Then optimize your selection. Which produces the most power.

Can an some one solve this problem for me so I can see how it is done?

 

[I like Serena]

I was asked the following question and had no idea how to solve it.

If we are concerned with cost, should we wire our christmas lights in series or parallel? Note you can also wire them in a parallel series. Then optimize your selection. Which produces the most power.

Can an some one solve this problem for me so I can see how it is done?

Suppose we wire 5 resistors of each 1 ohm in series.
And suppose we supply 5 V.
How much current will be generated?
How much power will be dissipated?

What if you wire them in parallel?

 

[Dustinsfl]

Suppose we wire 5 resistors of each 1 ohm in series.
And suppose we supply 5 V.
How much current will be generated?
How much power will be dissipated?

What if you wire them in parallel?

I figured out the first part. Suppose we have n resistors all with resistance R.

Series:
\(R_{\text{tot}} = nR\) so \(I_s = \frac{V}{nR}\). Therefore, \(P_s = \frac{V^2}{nR}\)

Parallel:
\(\frac{1}{R_{\text{tot}}} = \frac{n}{R}\) so \(I_p = \frac{nV}{R}\) and \(P_p = \frac{nV^2}{R}\)

For simplicity, we can assume the power is generated hourly. In parallel, we have more watts per hour and electricity is priced in kilowatts per hour so the total cost is greater in parallel.

The solution is then to wire in parallel series. Now the question is how do I setup the most efficient parallel series and show that is the case? It appears to be a Caculus optimization problem but I am not sure on how to do it.

 

[I like Serena]

I figured out the first part. Suppose we have n resistors all with resistance R.

Series:
\(R_{\text{tot}} = nR\) so \(I_s = \frac{V}{nR}\). Therefore, \(P_s = \frac{V^2}{nR}\)

Parallel:
\(\frac{1}{R_{\text{tot}}} = \frac{n}{R}\) so \(I_p = \frac{nV}{R}\) and \(P_p = \frac{nV^2}{R}\)

For simplicity, we can assume the power is generated hourly. In parallel, we have more watts per hour and electricity is priced in kilowatts per hour so the total cost is greater in parallel.

Exactly.

The solution is then to wire in parallel series.

There appear to be 2 questions.
For the lowest cost they need to be in series.
For the highest power output they need to be in parallel.

Now the question is how do I setup the most efficient parallel series and show that is the case? It appears to be a Caculus optimization problem but I am not sure on how to do it.

Putting any 2 resistors in series reduces the current and therefore the power.
So for maximum power the optimal solution is to put them all in parallel.

 

[Dustinsfl]

There appear to be 2 questions.
For the lowest cost they need to be in series.
For the highest power output they need to be in parallel.

Putting any 2 resistors in series reduces the current and therefore the power.
So for maximum power the optimal solution is to put them all in parallel.

I understand this but the solution is to wire the lights in a circuit with parallel and series. The question is then how to we optimize this wiring so cost isn't outrageous but if one light goes out the whole chain isn't broken.

 

[I like Serena]

I understand this but the solution is to wire the lights in a circuit with parallel and series. The question is then how to we optimize this wiring so cost isn't outrageous but if one light goes out the whole chain isn't broken.

It appears you have just introduced a new condition: if one light goes out the whole chain isn't broken.

How do you think they might be wired for lowest cost?

 

[Dustinsfl]

It appears you have just introduced a new condition: if one light goes out the whole chain isn't broken.

How do you think they might be wired for lowest cost?

The lowest cost would be to put a parallel at the n/2 location but would that be over all optimal? I understand it keeps cost down but if n is sufficently large a light out would still take down many lights. I was hoping there was a caculus way to figure out on how to determine parallel locations and how many lights in parallel.

 

[I like Serena]

The lowest cost would be to put a parallel at the n/2 location but would that be over all optimal? I understand it keeps cost down but if n is sufficently large a light out would still take down many lights. I was hoping there was a caculus way to figure out on how to determine parallel locations and how many lights in parallel.

If no light is allowed to go out when 1 breaks, that means each light must have an alternate path before and after to ensure current can always flow.
This is achieved by putting each light in parallel with 1 other light.
And then putting all these combinations in series.

I am not aware of a calculus way to solve a problem like this.
It's more of a graph theory question.